Maximizing volume of a box using differentiation

The net for this box would be:

Net for a box

The volume of the box is V = x2y

We are told that the surface area of the box is 64 cm2. The area of the base of the box is x2 and the area of each side is xy, so the area of the base plus the area of the 4 sides is given by:

x2 + 4xy = 64 cm2

Solving for y gives:

`y=(64-x^2)/(4x)=16/x-x/4`

So the volume can be rewritten:

`V=x^2y`

`=x^2(16/x-x/4)`

`=16x-x^3/4`

Now

`(dV)/(dx)=16-(3x^2)/4`

and this is zero when

`x=+- 8/sqrt(3) ~~ 4.62`

(Note: The negative case has no practical meaning.)

Is it a maximum?

`(d^2V)/(dx^2)=-(3x)/2`

and this is negative when x is positive. So it is a MAX.

So the dimensions of the box are:

Base 4.62 cm × 4.62 cm and sides 2.31 cm.

The maximum possible volume is

V = 4.62 × 4.62 × 2.31 ≈ 49.3 cm3

Check: Area of material:

x2 + 4xy = 21.3 + 4 × 4.62 × 2.31 = 64

Checks OK.

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